作业帮由数列前n项和,怎么用mathematica求通项公式?比如湖南高考数学第十五题http://wenku.baidu.com/link?url=T9ZfaHOG8SAJTyX9-eZq5kftOq8NGqMo9v9XV9s9djdsbFElydxyCTNSqdn53PqJYSx4Y6-FURpkaq_DXuCsx8YFnvMMONhkXnFw5p0OmA7求解,要用mathem
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/16 08:21:05
由数列前n项和,怎么用mathematica求通项公式?比如湖南高考数学第十五题http://wenku.baidu.com/link?url=T9ZfaHOG8SAJTyX9-eZq5kftOq8NGqMo9v9XV9s9djdsbFElydxyCTNSqdn53PqJYSx4Y6-FURpkaq_DXuCsx8YFnvMMONhkXnFw5p0OmA7求解,要用mathem
由数列前n项和,怎么用mathematica求通项公式?
比如湖南高考数学第十五题http://wenku.baidu.com/link?url=T9ZfaHOG8SAJTyX9-eZq5kftOq8NGqMo9v9XV9s9djdsbFElydxyCTNSqdn53PqJYSx4Y6-FURpkaq_DXuCsx8YFnvMMONhkXnFw5p0OmA7
求解,要用mathematica!
由数列前n项和,怎么用mathematica求通项公式?比如湖南高考数学第十五题http://wenku.baidu.com/link?url=T9ZfaHOG8SAJTyX9-eZq5kftOq8NGqMo9v9XV9s9djdsbFElydxyCTNSqdn53PqJYSx4Y6-FURpkaq_DXuCsx8YFnvMMONhkXnFw5p0OmA7求解,要用mathem
其实分奇偶讨论就好.
直接求试过,解不出.
Clear["Global`*"];
Simplify[af[2 n] == (-1)^(2 n) (af[2 n] + af[2 n - 1]) + 1/2^(2 n),
Assumptions -> n \[Element] Integers]
af[2 n - 1] == - 1/2^(2 n)
Simplify[- 1/2^(2 n + 2) == -(- 1/2^(2 n + 2) + af[2 n]) +
1/2^(2 n + 1), Assumptions -> n \[Element] Integers]
sexp = sf[n] == (-1)^n (sf[n] - sf[n - 1]) - 1/2^n
In[11]:= Simplify[sexp /. n -> 2 k,
Assumptions -> k \[Element] Integers]
Out[11]= 1 + 4^k sf[-1 + 2 k] == 0
Simplify[sexp /. n -> 2 k + 1,
Assumptions -> k \[Element] Integers](*Esc el Esc*)
2^(-1 - 2 k) + 2 sf[1 + 2 k] == sf[2 k]
In[16]:= Sum[-2^(-2 k), {k, 1, m},
Assumptions -> m \[Element] Integers]
Out[16]= -(1/3) 2^(-2 m) (-1 + 2^(2 m))
-(1/3) 2^(-100) (-1 + 2^100))